Question

Statement-$1$: $\overrightarrow{l}=a\overline{i}+b\overline{j}+c\overline{k},\overrightarrow{m}=b\overline{i}+c\overline{j}+a\overline{k}$, $\overrightarrow{n}=c\overline{i}+a\overline{j}+b\overline{k}$ are coplanar (where $a,b,c$ are positive) then $a=b=c$.

Statement-$2$: If $l\overrightarrow{a}+m\overrightarrow{b}+n\overrightarrow{c}=\overrightarrow{0}$ such that $l,m,n$ not all zero then $\overrightarrow{a}$, $\overrightarrow{b}$ $\overrightarrow{c}$ are coplanar.

• A. Both Statement $1$ and Statement $2$ are true and Statement $2$ is the correct explanation of Statement $1$.
• B. Both Statement $1$ and Statement $2$ are true but Statement $2$ is not correct explanation of Statement $1$.
• C. Statement $1$ is true, Statement $2$ is false.
• D. Statement $1$ is false, Statement $2$ is true.

Answer 1

The correct option is B Both Statement $1$ and Statement $2$ are true but Statement $2$ is not correct explanation of Statement $1$.

Since, $\overrightarrow{l},\overrightarrow{m},\overrightarrow{n}$ are coplanar, we have

$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$

This gives $a^3+b^3+c^3=3abc$
Which implies,

$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$

Since, $a,b,c$ are all positive, $a+b+c\ne 0$

$\therefore a^2+b^2+c^2-ab-bc-ca=0$

$\therefore (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$

Thus, $a=b,b=c,c=a$

Statement $2$ is always true but it does not explain statement $1$.

Hence, option $B.$